Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / O (Statistics)

Assume independent survival times and positive rate $\lambda$. With $S={\sum }_i{t}_i$, the likelihood is $L(\lambda )={\lambda }^n{e}^{-\lambda S}$ on the observed nonnegative data. The score and negative expected second derivative are $n/\lambda -S$ and $n/{\lambda }^2$. For $S>0$, which holds almost surely under the model,The usual independent-sample likelihood asymptotics give $\sqrt{n}(\hat{\lambda }-\lambda )\Rightarrow N(0,{\lambda }^2)$. Thus an approximate Wald interval isFor a positive interval with the same first-order justification, apply the delta method to $\log \hat{\lambda }$, obtaining endpoints $\hat{\lambda }\exp (\pm z_{1-\alpha /2}/\sqrt{n})$.

A single individual survives until $T$ with probability $s=e^{-\lambda T}$. Thus $N_T\sim \mathrm{Bin}(n,s)$ and the new likelihood is $\left(\genfrac{}{}{0ex}{}{n}{n_T}\right)e^{-\lambda T{n}_T}(1-e^{-\lambda T}{)}^{n-n_T}$. For $0<n_T<n$, its maximum occurs at $s=n_T/n$, givingIf $n_T=0$, the supremum occurs as $\lambda \to \infty$ and there is no finite maximum. If $n_T=n$, it occurs at the boundary $\lambda =0$ of the extended parameter space, or is only a supremum when the rate is required to be strictly positive.

A binomial experiment has information $n/[s(1-s)]$ for $s$. Since $ds/d\lambda =-Ts$, parameter transformation givesAs quantified by Fisher information in an exponential survival snapshot, the ratio of this information to the complete-data information is $z^2/(e^z-1)$ with $z=\lambda T>0$, strictly less than one by the supplied inequality. Hence every fixed survival snapshot loses information. Under regular large-sample conditions its efficient rate estimator has larger asymptotic variance, and correspondingly wider uncertainty intervals, than the estimator based on all individual survival times. This comparison concerns information and asymptotic uncertainty, not a claim that every realized interval in one experiment is wider.

For a sample size $m\ge 2$ define $\overline{X}=m^{-1}{\sum }_i{X}_i$ and $S_X^2=(m-1{)}^{-1}{\sum }_i(X_i-\overline{X}{)}^2$. The normal-sample theorem givesIt also states that the mean and sample variance are independent. Combining (i) and (ii), and the independence across samples, the correctly centred pivot isThe source reverses the mean difference inside the subtraction. Its literal statistic is generally noncentral when ${\mu }_X\ne {\mu }_Y$, rather than having a central Student distribution. Under the null hypothesis both differences are zero, so this misprint does not change the requested null test.

For a two-sided level-$\alpha$ test reject when $|T|>t_{m+n-2,1-\alpha /2}$. In the numerical example the pooled variance is $0.038$, the standard error is $\sqrt{0.038(2/5)}\approx 0.12329$, and $T\approx -2.190$. There are eight degrees of freedom and the supplied critical value is $2.31$. Thus do not reject equal mean cell volume at the 5% level. This does not establish equality of the means; it states that the observed difference does not cross this test's two-sided threshold.

Age matching introduces a shared block factor: the two observations in a pair can be associated, and their age-dependent means also undermine the simple unblocked independent-samples model. A paired t-test on the within-pair drug-minus-placebo differences accounts for that design, assuming independent pairs and the corresponding normal-difference model. Shared age variation cancels in those differences; positive within-pair association reduces their variance and can increase power, although the reduction in degrees of freedom means improvement is not automatic. The paired test cannot be computed from the two marginal means and variances alone: the within-pair differences or covariance are needed.

Independence makes a difference of the two normal sample means normal, with means subtracting and variances adding. Therefore

The variables $(m-1)S_X^2/{\sigma }^2$ and $(n-1)S_Y^2/{\sigma }^2$ are independent chi-squared variables with $m-1$ and $n-1$ degrees of freedom. Independence follows because the two whole samples are independent. Their sum is chi-squared with the sum of the degrees of freedom, soWithin each normal sample the sample mean is independent of its sample variance, and independence of the samples makes the pair of means jointly independent of the pair of variances. Hence this pooled chi-squared variable is independent of the standardized mean difference, which is the independence required for the Student pivot.